Mathematics

Evaluate:
$$\displaystyle\int{{x}^{2}{(3-x)}^{5}dx}$$


SOLUTION
$$\displaystyle =\int x^2 (3-x)^5dx$$
        $$ I$$       $$II$$

$$\displaystyle =x^2 \int(3-x)^5dx - \int\left(\dfrac{d(x^2)}{dx} \int (3-x)^5dx\right)dx$$

$$\displaystyle = x^2 \dfrac{(3-x)^6}{6} (-1) - \int 2x \dfrac{(3-x)^6}{6} (-1)dy$$

$$\displaystyle = \dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3} \int x(3-x)^6dy$$
                                           $$I$$     $$II$$

$$\displaystyle = \dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3} \left[ x\int (3-x)^6 dx - \int\left(\dfrac{dx}{dx}\int (3-x)^6 dx\right)dx\right]$$

$$\displaystyle =\int \dfrac{-x^2}{6}(3-x)^6 + \dfrac{1}{3} \left[x \dfrac{(3-x)^7}{7} (-1)-\int \dfrac{(3-x)^7}{7}(-1)dx\right]$$ 

$$\displaystyle = \dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3} \left[-\dfrac{x}{7}(3-x)^7 + \dfrac{1}{7}\int (3-x)^7dx\right]$$

$$\displaystyle =-\dfrac{x^2}{6} (3-x)^6 + \dfrac{1}{3} \left[-\dfrac{x}{7} (3-x)^7 + \dfrac{1}{7} \dfrac{(3-x)^8}{8} (-1) \right]+C$$

$$\displaystyle =\dfrac{-x^2}{6} (3-x)^6 + \dfrac{1}{3}\left[ -\dfrac{x}{7} (3-x)^7 - \dfrac{1}{56} (3-x)^8 \right]+ C$$
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Subjective Medium Published on 17th 09, 2020
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